LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-07

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Finite roof squares and composable pairs can be cleared

Statement

Let $f : X \to Y$ and $f^{'} : X^{'} \to Y^{'}$ be ordinary arrows, and let $α : Q X \to Q X^{'}$ , $β : Q Y \to Q Y^{'}$ satisfy $β Q f = Q f^{'} α$ . There exist $f^{''} : X^{''} \to Y^{''}$ , $k : X \to X^{''}$ , $l : Y \to Y^{''}$ , and denominators $s : X^{'} \to X^{''}$ , $t : Y^{'} \to Y^{''}$ , such that $f^{''} k = l f$ , $f^{''} s = t f^{'}$ , $α = Q (s)^{- 1} Q (k)$ , $β = Q (t)^{- 1} Q (l)$ . Moreover two composable localized arrows and their composite can simultaneously be represented by ordinary arrows after denominator isomorphisms of the three objects.

Facts & Assumptions

Given: Let $f : X \to Y$ and $f^{'} : X^{'} \to Y^{'}$ be ordinary arrows, and let $α : Q X \to Q X^{'}$ , $β : Q Y \to Q Y^{'}$ satisfy $β Q f = Q f^{'} α$ . There exist $f^{''} : X^{''} \to Y^{''}$ , $k : X \to X^{''}$ , $l : Y \to Y^{''}$ , and denominators $s : X^{'} \to X^{''}$ , $t : Y^{'} \to Y^{''}$ , such that $f^{''} k = l f$ , $f^{''} s = t f^{'}$ , $α = Q (s)^{- 1} Q (k)$ , $β = Q (t)^{- 1} Q (l)$ . Moreover two composable localized arrows and their composite can simultaneously be represented by ordinary arrows after denominator isomorphisms of the three objects.

[F1]

Every localized arrow has either roof orientation, and equality of ordinary arrows is detected by a denominator (The calculus of fractions constructs the localization).

[F2]

Composition of roof classes is independent of representatives and is associative (Composition of roofs is well defined).

Proof

1.1

Write $α = Q (s)^{- 1} Q (k)$ . The outgoing Ore square for $s, f^{'}$ gives $t : Y^{'} \to Y^{''}$ in $S$ and $f^{''} : X^{''} \to Y^{''}$ with $f^{''} s = t f^{'}$ . Write $β = Q (q)^{- 1} Q (i)$ , and apply Ore to $t, q$ to find $r : Y^{''} \to Y^{(3)}$ in $S$ and $j$ with $r t = j q$ . Replace $f^{''}, t$ by $r f^{''}, r t$ and put $l = j i$ . Then $β = Q (t)^{- 1} Q (l)$ and the right square commutes.

F1F2

2.1

The localized commuting square now gives $Q (l f) = Q (f^{''} k)$ . Dual equality detection supplies $d$ in $S$ with $d l f = d f^{''} k$ . Replace $l, f^{''}, t$ by $d l, d f^{''}, d t$ ; both ordinary squares now commute and $d t \in S$ . This proves the square assertion, including identity or coincident arrows.

F1step 1.1algebra

3.1

For a composable pair $α : Q X \to Q Y$ , $β : Q Y \to Q Z$ , write $α = Q (s)^{- 1} Q (f)$ with $s : Y \to Y^{'}$ in $S$ . Write $β Q (s)^{- 1} = Q (t)^{- 1} Q (g)$ with $t : Z \to Z^{'}$ in $S$ and $g : Y^{'} \to Z^{'}$ . Thus after the object comparisons $1_{X}, Q (s), Q (t)$ the pair is $Q (f), Q (g)$ and its composite is $Q (g f)$ . The equalities follow from the proved composition law, not from an assumption about arbitrary diagrams.

F1F2algebra∎

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Sources

13.5.5–13.5.6, including all TR1–TR4 proof paragraphs (standard reference, not scraped)
Lemma 4.27.10, complete proof including footnote common denominator construction (standard reference, not scraped)